Chained Crop Loss
A 4:3 frame converted to 16:9 and then to 9:16 keeps 23.7% of its area, so 76.3% is lost. Adding the two published losses gives 93.4%, which overstates the damage by 17.1 points. Cropping 4:3 straight to 9:16 loses 57.8%, so the middle step costs 18.5 extra points.
Direct answer: 76.3% lost, not 93.4%Two ways to get the number wrong
Both mistakes come from the same place: treating a sequence of crops as if it were one event.
Adding the steps. 4:3 to 16:9 removes 25.00% and 16:9 to 9:16 removes 68.36%, so the pair looks like 93.36%. It cannot be, because the second crop never sees the pixels the first one already threw away. The second loss applies to what survived, not to the original frame.
Quoting the straight crop. Cropping 4:3 directly to 9:16 removes 57.81%. That is the correct number for a single operation, and the wrong number for a workflow that exports a 16:9 file in between. If a 16:9 master or a 16:9 delivery exists anywhere in the chain, the frame has already paid twice.
The real figure for the three-step chain is 76.27% by the formula and 76.25% by an integer pixel pass. Both sit between the two wrong answers, which is why the mistake is easy to miss: one error makes the problem look fatal, the other makes it look mild.
Why the losses multiply
One crop from ratio rs to ratio rt keeps min(rs, rt) ÷ max(rs, rt) of the area. Each crop then keeps that fraction of whatever is left, so a chain keeps the product of the per-step fractions.
Chain kept = k1 × k2 × … × kn, where ki = min(ri−1, ri) ÷ max(ri−1, ri)
Chain loss = 1 − (k1 × k2 × … × kn)
Worked example on a 4:3 master at 1920 × 1440, which is 2,764,800 pixels. Cropping to 16:9 keeps the full width and cuts the height to 1920 × 9 ÷ 16 = 1080, so 1920 × 1080 = 2,073,600 pixels remain, or 75.00%. Cropping that to 9:16 keeps the height and rebuilds the width as 1080 × 9 ÷ 16 = 607.5, rounded to 608, so 608 × 1080 = 656,640 pixels remain.
The formula path gives 0.75 × 0.316406 = 0.237305, or 76.27% lost. The pixel path gives 23.75% kept, or 76.25% lost. The 0.02-point gap is the half pixel in 607.5 that has to become a whole 608.
When an extra step is free
Not every intermediate ratio costs something. Run all 720 ordered triples of the ten ratios on this site and the split is clean:
| Middle ratio | Triples | Extra loss vs a straight crop |
|---|---|---|
| Between source and target | 240 of 720 | 0.00 points in every case |
| Outside that interval | 480 of 720 | at least 1.13 points, up to 91.96 |
If the middle ratio sits numerically between the source and the target, the two crops pass through it without extra damage, and the chain equals the straight crop exactly. If it sits outside, the frame is pulled past its destination and pulled back, and the difference is paid twice.
This explains a result that looks wrong at first glance. 4:3 → 16:9 → 1:1 → 9:16 loses 76.27%, the same as the three-step chain without 1:1, because 1:1 sits between 16:9 (1.778) and 9:16 (0.5625). A square normalising step is free. Going out to 21:9 in the middle is not.
Twelve chains, measured
Chain loss is the product rule. Sum of the steps is what you get by adding the published per-step losses. Straight crop is the single conversion from the first ratio to the last. The final column is the price of the intermediate step.
| Chain | Chain loss | Sum of the steps | Straight crop | Middle step costs |
|---|---|---|---|---|
| 4:3 → 16:9 → 9:16 | 76.27% | 93.36% | 57.81% | +18.46 pt |
| 4:3 → 1:1 → 9:16 | 57.81% | 68.75% | 57.81% | +0 pt |
| 16:9 → 4:3 → 9:16 | 68.36% | 82.81% | 68.36% | +0 pt |
| 4:3 → 9:16 → 16:9 | 86.65% | 126.17% | 25% | +61.65 pt |
| 4:3 → 16:9 → 1:1 → 9:16 | 76.27% | 112.5% | 57.81% | +18.46 pt |
| 16:9 → 21:9 → 9:16 | 81.63% | 99.7% | 68.36% | +13.27 pt |
| 3:2 → 16:9 → 9:16 | 73.3% | 83.98% | 62.5% | +10.8 pt |
| 4:3 → 3:2 → 16:9 | 25% | 26.74% | 25% | +0 pt |
| 1:1 → 16:9 → 9:16 | 82.2% | 112.11% | 43.75% | +38.45 pt |
| 4:3 → 16:9 → 2.39:1 | 44.21% | 50.62% | 44.21% | +0 pt |
| 16:9 → 9:16 → 4:5 | 77.75% | 98.05% | 55% | +22.75 pt |
| 4:3 → 5:4 → 9:16 | 57.81% | 61.25% | 57.81% | +0 pt |
Rows with a zero in the last column are the monotone cases: 4:3 → 1:1 → 9:16, 16:9 → 4:3 → 9:16, 4:3 → 3:2 → 16:9, 4:3 → 16:9 → 2.39:1 and 4:3 → 5:4 → 9:16 all cost exactly what the straight crop costs. The 1:1 row is the clearest example: a square step between a landscape and a portrait ratio never adds damage.
The worst kind of middle step
21:9 to 2.39:1 is the cheapest conversion on this site at 2.37%, or 2.4% on the one-decimal table. Route it through 9:16 and the same endpoint costs 94.33%.
The two wide ratios are 2.4% apart, so almost nothing needs to move. A portrait stop in the middle forces the frame down to 0.5625 and back up to 2.39, and both moves discard most of the width. The detour is longer than the trip.
What this changes in practice
Pick the master ratio for the chain, not for one delivery. If a vertical cut is coming, a 4:3 or 1:1 master reaches 9:16 at a lower price than a 16:9 master does: 57.81% lost against 68.36%. The 16:9 master only wins if 16:9 is the final output and nothing else follows.
Keep intermediate exports inside the interval. When a proxy or a delivery file has to exist between the master and the final crop, give it a ratio between the two. A 16:9 proxy between 4:3 and 9:16 costs 18.46 points; a 1:1 proxy costs nothing.
Count the steps you actually take. The difference between 93.36% and 57.81% is not a rounding issue, it is a decision about which files exist. The calculator above reports all three numbers at once so the cost of each file is visible before it is made.
Chained crop questions
How much is lost converting 4:3 to 9:16 through 16:9?
76.27% of the area, leaving 23.73%. On a 1920 × 1440 master the surviving frame is 608 × 1080, which is 23.75% of the original pixels.
Why is the total not 93.4%?
Because the second crop takes a share of what the first one left, not a share of the original frame. 25.00% of the frame goes first, then 68.36% of the remaining 75.00%, which is 51.27% of the original. The two together remove 76.27%.
Does the order of two middle steps matter?
No. The kept fractions multiply, so swapping two intermediates gives the same result. Changing which ratio is in the middle does matter: it decides whether the chain is monotone or makes an excursion.
Is a square proxy always free?
Yes, whenever one endpoint is landscape and the other is portrait, because 1:1 then sits between them. If both endpoints are on the same side of square, a 1:1 stop is an excursion and does cost something.
Does a higher resolution change these percentages?
No. Every figure here is a ratio of areas, so a 4K master and a 720p master lose the same share. Resolution changes the pixel counts, not the fractions.
How was this checked?
Two independent passes. The closed form multiplies min/max at each step; the second pass builds real integer crops step by step and counts surviving pixels. Across the twelve published chains the largest disagreement is 0.02 of a percentage point, from rounding to whole pixels. The 720-triple rule was evaluated on the same closed form and every triple was classified by whether the middle ratio falls inside the endpoint interval.
Read next
One conversion at a time: what a 16:9 to 9:16 crop costs. The same fraction read from the other side: crop loss and letterbox coverage are one number. And why 21:9 never lands on whole pixels.
Method note
No outside data is used on this page. Every figure is arithmetic on the definition of a centred crop, so any cell can be reproduced with a calculator and the two formulas above. The formula, the integer-crop verification, the twelve chains, the 720-triple rule and the calculator are original work for this site.